Chapter 1 · Laboratory · Paper §2
Which cases survive?
Goal: find every permutation with neither forbidden shape. Section 3 turns the survivors into the corners of a polygon.
Each section has its own Undo edit and Redo edit buttons. On a keyboard, Ctrl+Z and Ctrl+Shift+Z undo and redo the last change in the section you are working in.
1. Six arrangements. Two forbidden shapes.
A permutation uses 1 to d once. Read three chosen positions from left to right. A pattern names their ranks: 1 = low, 2 = middle, 3 = high, rather than their actual values.
Each card graphs value (height) against position (left to right). Select a card to inspect it.
Change an arrangement. See every triple.
Numbers on dots are actual values. The horizontal axis labels positions.
Swap the values in two neighbouring positions:
All triples in this arrangement
2. Can we build every survivor without rejecting anything?
For four positions, start with the concrete example 4 | 2 3 | 1. A cut follows position 1 and position 3. The values rise inside each group; every value in a group is larger than every value in the groups to its right. These groups are called increasing blocks.
Choose cuts between positions
The numbered boxes are positions to fill. Toggle a cut after a box.
Fill the groups with values
From left to right: take the largest unused values needed for the group, and put them in increasing order.
Every cut choice, next to its arrangement
The general result — Lemma 2.1. The survivors are exactly these increasing blocks with larger values before smaller blocks. There are 2^(d−1) of them: each of the d−1 spaces independently has a cut or has no cut.
Why does this describe every survivor?
Locate the largest value, d. Everything before it must rise: a drop followed by d would be a 213. Everything before d must exceed everything after it: an increasing cross-pair together with d would be a 132. So the first group contains the largest available values in increasing order. Repeat on the remaining suffix.
Conversely, three values from these blocks can only have orders 123, 231, 312 or 321. A drop determines each cut uniquely. This gives a bijection for every d; the finite galleries above illustrate it.
3. Four arrangements make a polygon. What else lies inside?
For d = 3, treat each surviving permutation as a point. The polytope is their convex hull: all their weighted averages with nonnegative weights summing to one. Since the three values always sum to 6, two coordinates suffice.
Use prefix deficits: subtract the first value from its largest possible value, 3; subtract the sum of the first two from its largest possible sum, 5. The table shows these subtractions, and the way back, for the selected point.
Axes: deficits u₁, u₂. Labels: cut set · arrangement ({1} = a cut after position 1, ∅ = no cuts). Large wine discs: the vertices, m times a survivor’s deficits. Filled blue discs: interior lattice points. Hollow rings: other boundary points. Dark ring: the selected point.
Example, then the general result. Here d = 3, and every lattice point for m ≤ 3 is counted by brute force: an example, not a proof.
For every d the paper writes the deficits as u_k = M_k − S_k(x), where S_k(x) is the sum of the first k values and M_k its largest possible value (eq. (4), p. 5). The polytope becomes Q_d = {u : u_k ≥ 0, 2u_k − u_(k−1) − u_(k+1) ≤ 2} (eq. (6), with u_0 = u_d = 0); Figure 1(a) draws Q_3.
The way back, eq. (5), has integer coefficients and the deficit map is unimodular (p. 8), so lattice points match one to one at every dilation. The counts 1, 5, 13, 25 at m = 0, 1, 2, 3 begin the series of eq. (1); Table 1 (p. 2) gives its numerator H_3 = (1+t)^2.
List every counted lattice point and the inequalities used
For d = 3, eq. (6) reads u₁,u₂ ≥ 0, 2u₁−u₂ ≤ 2 and 2u₂−u₁ ≤ 2. At dilation m, replace both right sides by 2m. The program tests all integer pairs in the bounded square 0 ≤ u₁,u₂ ≤ 2m.
Eq. (5) at dilation m gives the way back: x₁=3m−u₁, x₂=2m+u₁−u₂, and x₃=m+u₂. Integer coordinates remain integer in both directions (the map is unimodular, p. 8), so the deficit polygon and the original permutation polytope have the same lattice-point counts. Each entry below shows both coordinates.
Counting these points at m = 0, 1, 2, 3 gives 1, 5, 13, 25. Two open explorations continue this count: One point, two views compares the five points of m = 1 in both coordinate systems, and Counts to a polynomial turns 1, 5, 13, 25 into a polynomial and the numerator H₃(t) without counting a dot twice.
Next: these cut sets make a cube
For d = 3 the four survivors are the corners of this quadrilateral. For every d, the 2^(d−1) survivors are the corners of a polytope whose faces fit together like those of a (d−1)-dimensional cube. Counting the corners does not show that. Three open explorations follow the same cut sets, starting from this polygon:
What does this prepare us to study?
Every later chapter is an open exploration; start anywhere below.
Scope of the paper’s result on zeros. H₁=H₂=1 have no zeros. For 3 ≤ d ≤ 1000 an exact certificate establishes negative real zeros (Theorem 1.3, computed on the author’s table, which is not bundled here); an analytic proof covers d ≥ 2⁷² (Theorem 1.4). The interval 1001 ≤ d < 2⁷² remains open in the arXiv paper. This demo bundles an independent reconstruction of the table only up to d = 30, and our small enumerations prove no assertion about that interval.